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I am faced with the problem of calculating the eigenfunctions for an operator of the form:

$(Kf)(x) = \int_{-\infty}^{\infty} K(x-\alpha y) f(y)dy $

Does anyone know for which functions (or types of functions) K analytic solutions for the eigenfunctions or eigenvalues are known? Or are there methods for solving this in particular cases?

After user1952009's comment I managed to make the following progress: using a Fourier transform on the equation the following form of a Fourier space eigenvalue equation can be obtained:

$ \hat{K}(\omega) \hat{f}(\alpha \omega) = \lambda \hat{f}(\omega) $

where $\hat{K}(\omega)$ is some function and $\lambda$ is the eigenvalue.

Does anyone know how to solve this equivalent equation?

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    $\begingroup$ the case $\alpha = \pm 1$ is much simpler : as every convolution operator, it is diagonal in the Fourier basis. for $\alpha \ne \pm 1$, you get as the eigenvector equation $f(x/a) \ast h(x) = \lambda f(x)$ which means that $\hat{f}(\omega) = \lambda \hat{f}(a \omega)$ i.e. $\hat{f}(\omega) = |\omega|^{\ln \lambda / \ln a}$ or something like that $\endgroup$ – reuns Jan 29 '16 at 20:46
  • $\begingroup$ @user1952009 Thank you that was already very helpful (the comment kind of deserves to be an answer). I've tried it and what I'm not sure on is how to obtain the eigenvalue spectrum from this? also did you miss out h in the Fourier equation? $\endgroup$ – Wolpertinger Jan 29 '16 at 23:36

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